# Mathematical relationship between two line segments perpendicular to a circle

Hello,

What is the mathematical relationship between the length of two line segments, each of which extends radially (perpendicularly) from a separate point on a circle to a line which is tangent to a third point on the circle?

For example, the following diagram shows a circle with origin o and radius r (not = 1) which is centered on the intersection of horizontal axis x and vertical axis y. Line z is tangent to the circle at point p which is the top intersection of the circle and vertical axis y. Line segments s & t (red & green respectively) each begin on the circle and extend radially to line x. Angle a is formed by the vertical axis y and the radial extension of line segment s while angle b is formed by the radial extensions of line segments s & t. Arc d extends from the origin of line segment s to the origin of line segment t.

Questions:

1) What is the mathematical relationship between the length of line segments s & t?

2) What is the length of arc d (may be expressed in radians)?

Thanks
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Commented:
1)
s=(secant(a)-1)*r
t = (secant(a+b)-1)*r
2)
d = r*b
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If it helps you to think about this from first principles, you have two right angle triangles Y, o, s and Y, o, t with a third triangle embedded s, t and a segment of the tangent. Since two of the triangles are right angled, you can work out all the numbers from first principles if you wish. It comes out to #1 in the above post.

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Author Commented:
Thanks for the response.

Could you show the derivation please?

I understand that

but I have not been able to determine how you arrived at that solution.
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Commented:
hypotenuse = r+s or r+t
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Commented:
>> 1. What is the mathematical relationship between the length of line segments s & t?
To answer this equation, you should express s and t in terms of each other.

Using ozo's equations:
s=(secant(a)-1)*r
t = (secant(a+b)-1)*r

Easiest way to express one in terms of the other is to divide lhs and rhs of the two equations:

s/t = { (secant(a)-1)*r } / { (secant(a+b)-1)*r }

You can cancel the r factor in the numerator and denominator leaving:

s/t = (secant(a)-1) / (secant(a+b)-1)

This means that for a constant pair of angles, a, b, the ratio of s/t is unaffected by the size of the circle since the r factor has been removed.

You can now write:
s = t * (secant(a)-1) / (secant(a+b)-1)

====================

Just to make sure that you understand how to derive ozo's equations:

adjacent in both right triangles is just the vertical line, OP, which is just the radius of the circle. So, OP = r.

hyp associated with angle a is r+s
hyp associated with angle (a+b) is r+t

sec(a)     = (r+s)/r ==> (r+s) = r sec(a)     ==> s = r sec(a) - r = r( sec(a) - 1)
sec(a+b) = (r+t)/r ==> (r+t) = r sec(a+b) ==> t = r sec(a+b) - r = r( sec(a+b) - 1)
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